Chapter 13, Problem 22P

Find current I_o in the circuit of Fig. 13.91.

To determine

Calculate the current $I_o$ in the given circuit.

Answer to Problem 22P

The current $I_o$ is $\underset{\_}{3.129\angle 62.98°\text{A}}$.

Explanation of Solution

Given data:

Refer to Figure 13.91 in the textbook for the coupled coil circuit.

Calculation:

In Figure 13.91, replace the coupled inductor by dependent source model by using Figure 13.8. The modified circuit as shown in Figure 1.

In Figure 1, consider the followings.

$\begin{array}{l}{I}_{\text{a}}=I_1-I_3,\\ I_{\text{b}}=I_2-I_1,\\ I_{\text{c}}=I_3-I_2,\text{and}\\ I_o=I_3\end{array}$

From Figure 1, consider that the loops 1, 2 and 3 contain the currents $I_1$, $I_2$, and $I_3$ respectively.

Apply Kirchhoff's voltage law to the loop 1 in Figure 1.

$\begin{array}{l}\left[\begin{array}{l}j20I_{\text{c}}+j40\left(I_1-I_3\right)+j10I_{\text{b}}\\ -j30I_{\text{b}}+j80\left(I_1-I_2\right)-j10I_{\text{a}}\end{array}\right]=120\\ \left[\begin{array}{l}j20\left(I_3-I_2\right)+j40\left(I_1-I_3\right)+j10\left(I_2-I_1\right)\\ -j30\left(I_3-I_2\right)+j80\left(I_1-I_2\right)-j10\left(I_1-I_3\right)\end{array}\right]=120\end{array}$

$j100I_1-j60I_2-j40I_3=120$        (1)

Apply Kirchhoff's voltage law to the loop 2 in Figure 1.

$\begin{array}{l}\left[\begin{array}{l}j10I_{\text{a}}+j80\left(I_2-I_1\right)+j30I_{\text{c}}\\ -j30I_{\text{b}}+j60\left(I_2-I_3\right)-j20I_{\text{a}}+100I_2\end{array}\right]=0\\ \left[\begin{array}{l}j10\left(I_1-I_3\right)+j80\left(I_2-I_1\right)+j30\left(I_3-I_2\right)\\ -j30\left(I_2-I_1\right)+j60\left(I_2-I_3\right)-j20\left(I_1-I_3\right)+100I_2\end{array}\right]=0\end{array}$

$-j60I_1+\left(100+j80\right)I_2-j20I_3=0$        (2)

Apply Kirchhoff's voltage law to the loop 3 in Figure 1.

$\begin{array}{l}\left[\begin{array}{l}-j50I_3+j20I_{\text{a}}+j60\left(I_3-I_2\right)+j30I_{\text{b}}\\ -j10I_{\text{b}}+j40\left(I_3-I_1\right)-j20I_{\text{c}}\end{array}\right]=0\\ \left[\begin{array}{l}-j50I_3+j20\left(I_1-I_3\right)+j60\left(I_3-I_2\right)+j30\left(I_2-I_1\right)\\ -j10\left(I_2-I_1\right)+j40\left(I_3-I_1\right)-j20\left(I_3-I_2\right)\end{array}\right]=0\end{array}$

$-j40I_1-j20I_2+j10I_3=0$        (3)

Write equations (1), (2), and (3) in matrix form as follows.

$\left[\begin{array}{ccc}j100& -j60& -j40\\ -j60& \left(100+j80\right)& -j20\\ -j40& -j20& j10\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ I_3\end{array}\right]=\left[\begin{array}{c}120\\ 0\\ 0\end{array}\right]$        (4)

Write the MATLAB code to solve the equation (4).

A = [j*100 j*(-60) j*(-40); j*(-60) (100+j*80) j*(-20); j*(-40) j*(-20) j*10];

B = [120; 0; 0];

C = inv(A)*B

The output in command window:

C =

  0.45231 + 0.34154i

 -0.19385 + 0.71077i

  1.42154 + 2.78769i

From the MATLAB output, the currents $I_1$, $I_2$, and $I_3$ are,

$\begin{array}{c}{I}_1=\left(0.45231+j0.34154\right)\text{A}\\ =0.5668\angle 37.06°\text{A}\end{array}$

$\begin{array}{c}{I}_2=\left(-0.19385+j0.71077\right)\text{A}\\ =0.7367\angle 105.25°\text{A}\end{array}$

And

$\begin{array}{c}{I}_3=\left(1.42154+j2.78769\right)\text{A}\\ =3.129\angle 62.98°\text{A}\end{array}$

Write the expression for the current $I_o$ using Figure 1.

$I_o=I_3$

Substitute $3.129\angle 62.98°\text{A}$ for $I_3$.

$I_o=3.129\angle 62.98°\text{A}$

Conclusion:

Thus, the current $I_o$ is $\underset{\_}{3.129\angle 62.98°\text{A}}$.